University of Illinois Spring 2008
ECE 313: Solutions to Problem Set 5
1. (a) Xdenotes a binomial random variable with parameters (N, p). It counts the number of
occurrences of an event Aof probability pon Nindependent trials. Y=N− X counts
the number of occurrences of Ac, an event of probability 1 −p, on the Nindependent
trials and is thus a binomial random variable with parameters (N, 1−p).
(b) P{X is even}=N
0p0(1 −p)N+N
2p2(1 −p)N−2+N
4p4(1 −p)N−4+···
Now, from the binomial theorem (Ross, page 8) we get that
(x+y)N+ (−x+y)N= 2hN
0x0yN+N
2x2yN−2+N
4p4(1 −p)N−4+···i, and so,
setting x=p, y = 1 −p, we get P{X is even}=1
2(p+ 1 −p)N+ (−p+ 1 −p)N=
1
21 + (1 −2p)N.Notice that the probability is 1/2 for p= 1/2, 1 for p= 0, and 1 (or
0) for p= 1 according as Nis even or odd.
2. (a) P{Y is even}=P{Y = 0}+P{Y = 2}+P{Y = 4}+· ·· = exp(−λ)h1 + λ2
2! +λ4
4! +···i
= exp(−λ) cosh(λ) by recognizing the series in square brackets.
(b) 1
21 + (1 −2p)N=1
2h1 + 1−2λ
NNi→1
2[1 + exp(−2λ)] = exp(−λ)exp(λ)+exp(−λ)
2
= exp(−λ) cosh(λ).
(c) For fixed Y=k, the likelihood function (parameterized by θ) is L(θ;k) = θke−θ
k!.To
find the ML estimate of λ, we find ˆ
θthat maximizes L(θ;k). Setting the derivative of
L(θ;k) equal to zero, we get
∂L(θ;k)
∂θ =1
k!hkθk−1e−θ+θke−θ(−1)i=θk−1e−θ
k![k−θ]=0 ⇒ˆ
θML =k.
3. (a) On average, E[X] = 105 ×0.9 = 94.5 passengers show up for the flight.
(b)
P{X ≤ 100}= 1 −P{X >100}= 1 −
105
X
k=101
P{X =k}
= 1 −105
101(0.9)101(0.1)4−105
102(0.9)102(0.1)3−105
103(0.9)103(0.1)2
−105
104(0.9)104(0.1)1−105
105(0.9)105(0.1)0
= 1 −105
4(0.9)101(0.1)4−105
3(0.9)102(0.1)3−105
2(0.9)103(0.1)2
−105
1(0.9)104(0.1)1−105
0(0.9)105(0.1)0
= 0.9832 . . .
(c) We saw earlier that if Xis a binomial random variable with parameters (n, p), then
Y=n− X is a binomial random variable with parameters (n, 1−p).
(d) P{Y ≥ 5}= 1 −P{Y = 0} − P{Y = 1} − P{Y = 2} − P{Y = 3} − P{Y = 4}=
1−exp(−10.5) h1 + 10.5
1! +(10.5)2
2! +(10.5)3
3! +(10.5)4
4! i= 0.9789 . . . .
